没有比这更好的方法了。Bash只知道写入标准输出的状态码(整数)和字符串。

Bash模式返回标量和数组值对象:

定义

url_parse() { # parse 'url' into: 'url_host', 'url_port', ...
   local "$@" # inject caller 'url' argument in local scope
   local url_host="..." url_path="..." # calculate 'url_*' components
   declare -p ${!url_*} # return only 'url_*' object fields to the caller
}

调用

main() { # invoke url parser and inject 'url_*' results in local scope
   eval "$(url_parse url=http://host/path)" # parse 'url'
   echo "host=$url_host path=$url_path" # use 'url_*' components
}

你可以使用一个全局变量:

declare globalvar='some string'

string ()
{
  eval  "$1='some other string'"
} # ----------  end of function string  ----------

string globalvar

echo "'${globalvar}'"

这给了

'some other string'

没有比这更好的方法了。Bash只知道写入标准输出的状态码(整数)和字符串。

虽然有很多很好的答案，但它们都不是我想要的方式。下面是我的解决方案，要点如下:

帮助健忘的程序员

至少我会努力记住错误检查之后，像这样:

允许用换行字符\n赋值

有些解决方案不允许这样做，因为有些解决方案忘记了要赋值周围的单引号。正确的方法:eval "${returnVariable}='${value}'"或者更好:参见下面的下一点。

使用printf代替eval

只需尝试使用类似myFunction "date && var2"的东西来解决这里的一些假定的解决方案。Eval会执行给它的任何东西。我只想分配值，所以我使用printf -v "${returnVariable}" "%s" "${value}"代替。

对变量名冲突的封装和保护

如果一个不同的用户或至少是对函数了解较少的人(这可能是几个月后的我)正在使用myFunction，我不希望他们知道他必须使用全局返回值名称或禁止使用某些变量名称。这就是为什么我在myFunction的顶部添加了一个名称检查:

    if [[ "${1}" = "returnVariable" ]]; then
        echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
        echo "If that is still what you want to do please do that outside of myFunction()!"
        return 1
    fi

注意，如果你需要检查很多变量，这也可以放在函数本身中。 如果我仍然想使用相同的名称(这里:returnVariable)，我只是创建了一个缓冲变量，给myFunction，然后复制值returnVariable。

就是这样:

myFunction ():

myFunction() {
    if [[ "${1}" = "returnVariable" ]]; then
        echo "Cannot give the ouput to \"returnVariable\" as a variable with the same name is used in myFunction()!"
        echo "If that is still what you want to do please do that outside of myFunction()!"
        return 1
    fi
    if [[ "${1}" = "value" ]]; then
        echo "Cannot give the ouput to \"value\" as a variable with the same name is used in myFunction()!"
        echo "If that is still what you want to do please do that outside of myFunction()!"
        return 1
    fi
    local returnVariable="${1}"
    local value=$'===========\nHello World\n==========='
    echo "setting the returnVariable now..."
    printf -v "${returnVariable}" "%s" "${value}"
}

测试用例:

var1="I'm not greeting!"
myFunction var1
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var1:\n%s\n" "${var1}"

# Output:
# setting the returnVariable now...
# myFunction(): SUCCESS
# var1:
# ===========
# Hello World
# ===========

returnVariable="I'm not greeting!"
myFunction returnVariable
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "returnVariable:\n%s\n" "${returnVariable}"

# Output
# Cannot give the ouput to "returnVariable" as a variable with the same name is used in myFunction()!
# If that is still what you want to do please do that outside of myFunction()!
# myFunction(): FAILURE
# returnVariable:
# I'm not greeting!

var2="I'm not greeting!"
myFunction "date && var2"
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var2:\n%s\n" "${var2}"

# Output
# setting the returnVariable now...
# ...myFunction: line ..: printf: `date && var2': not a valid identifier
# myFunction(): FAILURE
# var2:
# I'm not greeting!

myFunction var3
[[ $? -eq 0 ]] && echo "myFunction(): SUCCESS" || echo "myFunction(): FAILURE"
printf "var3:\n%s\n" "${var3}"

# Output
# setting the returnVariable now...
# myFunction(): SUCCESS
# var3:
# ===========
# Hello World
# ===========

They key problem of any 'named output variable' scheme where the caller can pass in the variable name (whether using eval or declare -n) is inadvertent aliasing, i.e. name clashes: From an encapsulation point of view, it's awful to not be able to add or rename a local variable in a function without checking ALL the function's callers first to make sure they're not wanting to pass that same name as the output parameter. (Or in the other direction, I don't want to have to read the source of the function I'm calling just to make sure the output parameter I intend to use is not a local in that function.)

解决这个问题的唯一方法是使用一个单独的专用输出变量，比如REPLY(由ev1m4chine建议)，或者使用Ron Burk建议的约定。

但是，也可以让函数在内部使用固定的输出变量，然后在上面添加一些糖来对调用者隐藏这一事实，就像我在下面的示例中对call函数所做的那样。把这看作是概念的证明，但关键是

该函数总是将返回值赋给REPLY，也可以像往常一样返回退出码 从调用者的角度来看，返回值可以分配给任何变量(本地或全局)，包括REPLY(参见包装器示例)。函数的退出码是传递的，因此在if或while或类似结构中使用它们可以正常工作。 从语法上讲，函数调用仍然是一条简单的语句。

这样做的原因是调用函数本身没有局部变量，并且除了REPLY之外不使用其他变量，从而避免了任何名称冲突的可能性。当调用方定义的输出变量名被赋值时，我们实际上处于调用方的作用域(技术上讲，在调用函数的相同作用域中)，而不是在被调用函数的作用域中。

#!/bin/bash
function call() { # var=func [args ...]
  REPLY=; "${1#*=}" "${@:2}"; eval "${1%%=*}=\$REPLY; return $?"
}

function greet() {
  case "$1" in
    us) REPLY="hello";;
    nz) REPLY="kia ora";;
    *) return 123;;
  esac
}

function wrapper() {
  call REPLY=greet "$@"
}

function main() {
  local a b c d
  call a=greet us
  echo "a='$a' ($?)"
  call b=greet nz
  echo "b='$b' ($?)"
  call c=greet de
  echo "c='$c' ($?)"
  call d=wrapper us
  echo "d='$d' ($?)"
}
main

输出:

a='hello' (0)
b='kia ora' (0)
c='' (123)
d='hello' (0)